Rotational Motion MCQs for NEET — Physics Questions with Answers

In an elastic collision between two equal masses moving at right angles with equal speeds, the final velocity of the combined mass is v/sqrt(2) at an angle of 45° from the original directions. This result follows from the conservation of momentum and kinetic energy in the collision.

In an inelastic collision where a bullet of mass a and velocity b strikes a large block of mass c, the final velocity of the system is (a/(a+c))b. This result is derived from the conservation of momentum, considering the initial and final momenta of the system.

In an inelastic collision where the two bodies stick together, the total momentum before and after the collision must be conserved. Let the initial velocities be v and 0. By conservation of momentum, (mv + 2m0) = (m+2m)*v', where v' is the final velocity of the combined system. Solving this, we get v' = v/3.

Converting the given velocities to SI units, the initial velocity of the moving body is (3 km/h) = (3/3.6) m/s = 5/6 m/s. Using conservation of momentum, (m*(5/6) + 2m0) = 3mv', where v' is the final velocity of the combined mass. Solving this, we get v' = (5/18) m/s = 1 km/h.

Loss in K.E. = (initial K.E. – Final K.E.) of system

$\frac{1}{2}{m}_1{u}_1^2+\frac{1}{2}{m}_2{u}_2^2-\frac{1}{2}(m_1+m_2)V^2$

$=\frac{1}{2}3\times {\left(32\right)}^2+\frac{1}{2}\times 4\times {\left(5\right)}^2-\frac{1}{2}\times (3+4)\times {\left(5\right)}^2$ 

= 986.5 J

$v=36km\text{}/\text{}h=10m\text{}/\text{}s$

By law of conservation of momentum

$2\times 10=(2+3)V$ ⇒ V = 4 m/s

Loss in K.E. $=\frac{1}{2}\times 2\times {\left(10\right)}^2-\frac{1}{2}\times 5\times {\left(4\right)}^2=60J$  

Initial momentum = $\overrightarrow{P}=mv\hat{i}+mv\hat{j}$

$\left|\overrightarrow{P}\right|=\sqrt{2}mv$

Final momentum = 2m × V

By the law of conservation of momentum

$2m\times V=\sqrt{2}mv$ ⇒ $V=\frac{v}{\sqrt{2}}$

In the problem v = 10 m/s (given) 

∴ $V=\frac{10}{\sqrt{2}}=5\sqrt{2}m\text{}/\text{}s$

Because in perfectly inelastic collision the colliding bodies stick together and move with common velocity

According to law of conservation of momentum.

Momentum of neutron = Momentum of combination

⇒ $1.67\times {10}^{-27}\times {10}^8=(1.67\times {10}^{-27}+3.34\times {10}^{-27})v$

∴ $v=3.33\times {10}^{7}m\text{}/\text{}s$